MCAT Physics Question — New Force
- Nov 05, 2014
- MCAT Question of the Day
Two charges, Q1 and Q2, experience an attractive force between them, F1. If the distance between the charges is tripled, the valence on Q2 is changed, and the magnitude of the charge on Q1 increases by a factor of nine, what will the new force be between the charges?
The force between the charges is given by Coulomb’s law:
F = kQ1q2 / r2
We may start by noting that changing one of the charges from positive to negative (or vice versa), as the question indicates when it says the valence on one of the charges is flipped, would immediately change the force from attractive to repulsive, or vice versa. Thus, we may immediately narrow our choices down to (B) and (C).
Next, if the distance between the charges in tripled and one charge in increased by a factor of 9, we have the following:
F = k(9Q1)q2 / (3r)2
F = 9kQ1q2 / (9)(r2)
F = kQ1q2 / r2
Thus we are left with a magnitue of force that is unchanged. Therefore, the answer is (B). Same magnitude, opposite direction.
Want more MCAT practice?
We’ve got options for every schedule and learning style!
From the best online MCAT course created by top instructors with 524+ MCAT scores to the most representative full-length practice exams and private tutoring, we can custom tailor your MCAT prep to your goals!
Not sure which option is right for you? Schedule a free MCAT consultation with an MCAT expert using the form below. No obligation, just expert advice.
Search the Blog
Free ConsultationSchedule Now
Free MCAT Practice AccountNeed great MCAT practice?
Get the most representative MCAT practice possible when you sign up for our free MCAT Account, which includes a half-length diagnostic exam and one of our full-length MCAT practice exams.Learn More